1. **State the problem:** Find the maximum and minimum points of the function $$f(x) = x^3 + 3x^2 - 4x$$ and describe its graph. 2. **Find the derivative:** To find critical points, compute $$f'(x)$$: $$f'(x) = 3x^2 + 6x - 4$$ 3. **Set the derivative equal to zero:** Solve for $$x$$: $$3x^2 + 6x - 4 = 0$$ Divide both sides by 3: $$x^2 + 2x - \frac{4}{3} = 0$$ 4. **Use the quadratic formula:** For $$ax^2 + bx + c = 0$$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a=1$$, $$b=2$$, $$c=-\frac{4}{3}$$. Calculate the discriminant: $$\Delta = 2^2 - 4 \times 1 \times \left(-\frac{4}{3}\right) = 4 + \frac{16}{3} = \frac{28}{3}$$ 5. **Find the critical points:** $$x = \frac{-2 \pm \sqrt{\frac{28}{3}}}{2} = -1 \pm \frac{\sqrt{\frac{28}{3}}}{2}$$ Simplify the square root: $$\sqrt{\frac{28}{3}} = \frac{2\sqrt{21}}{3}$$ So, $$x = -1 \pm \frac{2\sqrt{21}}{3 \times 2} = -1 \pm \frac{\sqrt{21}}{3}$$ 6. **Calculate approximate values:** $$\sqrt{21} \approx 4.58$$ So critical points are approximately: $$x_1 = -1 + \frac{4.58}{3} = -1 + 1.53 = 0.53$$ $$x_2 = -1 - 1.53 = -2.53$$ 7. **Find the function values at critical points:** Calculate $$f(0.53)$$: $$f(0.53) = (0.53)^3 + 3(0.53)^2 - 4(0.53) \approx 0.15 + 0.84 - 2.12 = -1.13$$ Calculate $$f(-2.53)$$: $$f(-2.53) = (-2.53)^3 + 3(-2.53)^2 - 4(-2.53) \approx -16.17 + 19.20 + 10.12 = 13.15$$ 8. **Determine maxima and minima:** Use the second derivative: $$f''(x) = 6x + 6$$ Evaluate at $$x=0.53$$: $$f''(0.53) = 6(0.53) + 6 = 3.18 + 6 = 9.18 > 0$$ so this is a minimum. Evaluate at $$x=-2.53$$: $$f''(-2.53) = 6(-2.53) + 6 = -15.18 + 6 = -9.18 < 0$$ so this is a maximum. 9. **Summary:** - Maximum point at approximately $$(-2.53, 13.15)$$ - Minimum point at approximately $$(0.53, -1.13)$$ 10. **Graph description:** The cubic function has roots at $$x = -4, 0, 1$$. It rises to a maximum near $$x = -2.53$$, then falls to a minimum near $$x = 0.53$$, and then rises again. Final answer: Maximum point: $$\boxed{(-2.53, 13.15)}$$ Minimum point: $$\boxed{(0.53, -1.13)}$$